Find the sum of G.P. : 3, 6, 12, ……., 1536.

Common ratio = $\dfrac{6}{3}$ = 2.

Let 1536 be n^th term

$\therefore ar^{n - 1} = 1536 \\[1em] \Rightarrow 3.(2)^{n - 1} = 1536 \\[1em] \Rightarrow (2)^{n - 1} = 512 \\[1em] \Rightarrow (2)^{n - 1} = (2)^{9} \\[1em] \Rightarrow n - 1 = 9 \\[1em] \Rightarrow n = 10.$

Since, r > 1

$S = \dfrac{a(r^n - 1)}{(r - 1)} \\[1em] = \dfrac{3 \times (2^{10} - 1)}{2 - 1} \\[1em] = 3(2^{10} - 1) \\[1em] = 3(1024 - 1) \\[1em] = 3 \times 1023 \\[1em] = 3069.$

Hence, sum = 3069.